a can only be final here. Why? How can I reassign a in onClick() method without keeping it as private member? private void f(Button b, final int a){ b.addClickHandler(new ClickHandler() { @Override public void onClick(ClickEvent event) { int b = a*5; } }); } How can I return the 5 * a when it clicked? I mean, private void f(Button b, final int a){ b.addClickHandler(new ClickHandler() { @Override public void onClick(ClickEvent event) { int b = a*5; return b; // but return type is void } }); }


当前回答

有一个技巧允许匿名类在外部作用域中更新数据。

private void f(Button b, final int a) {
    final int[] res = new int[1];
    b.addClickHandler(new ClickHandler() {
        @Override
        public void onClick(ClickEvent event) {
            res[0] = a * 5;
        }
    });

    // But at this point handler is most likely not executed yet!
    // How should we now res[0] is ready?
}

然而,由于同步问题,这个技巧不是很好。如果处理程序稍后被调用,你需要1)同步访问res,如果处理程序从不同的线程调用2)需要有某种标志或指示res被更新

但是,如果匿名类在同一个线程中立即被调用,这个技巧就可以工作。如:

// ...

final int[] res = new int[1];
Runnable r = new Runnable() { public void run() { res[0] = 123; } };
r.run();
System.out.println(res[0]);

// ...

其他回答

匿名内部类中的方法可以在生成该类的线程终止后调用。在您的示例中,内部类将在事件分派线程上调用,而不是在创建它的线程中调用。因此,变量的作用域是不同的。所以为了保护这样的变量赋值范围问题,你必须声明它们为final。

试试这段代码,

创建数组列表并将值放入其中并返回:

private ArrayList f(Button b, final int a)
{
    final ArrayList al = new ArrayList();
    b.addClickHandler(new ClickHandler() {

         @Override
        public void onClick(ClickEvent event) {
             int b = a*5;
             al.add(b);
        }
    });
    return al;
}
private void f(Button b, final int a[]) {

    b.addClickHandler(new ClickHandler() {

        @Override
        public void onClick(ClickEvent event) {
            a[0] = a[0] * 5;

        }
    });
}

要理解这种限制的基本原理,请考虑以下程序:

public class Program {

    interface Interface {
        public void printInteger();
    }
    static Interface interfaceInstance = null;

    static void initialize(int val) {
        class Impl implements Interface {
            @Override
            public void printInteger() {
                System.out.println(val);
            }
        }
        interfaceInstance = new Impl();
    }

    public static void main(String[] args) {
        initialize(12345);
        interfaceInstance.printInteger();
    }
}

The interfaceInstance remains in memory after the initialize method returns, but the parameter val does not. The JVM can’t access a local variable outside its scope, so Java makes the subsequent call to printInteger work by copying the value of val to an implicit field of the same name within interfaceInstance. The interfaceInstance is said to have captured the value of the local parameter. If the parameter weren’t final (or effectively final) its value could change, becoming out of sync with the captured value, potentially causing unintuitive behavior.

由于Jon有实现细节的答案,另一个可能的答案是JVM不想处理已经结束他的激活的写入记录。

考虑这样一个用例,在这个用例中,你的lambdas不是被应用的,而是被存储在某个地方并稍后运行。

我记得在Smalltalk中,当你做这样的修改时,你会得到一个非法的商店。